Question

  

A 10.00-mL aliquot of 0.2000 M mercaptoacetic acid (HSCH₂CO₂H) is titrated with 0.1200 M KOH. Calculate the pH

a)after the addition of 0.00 mL KOH.

b)at the 1^st equivalence point.

c)after the addition of 5.00 mL of KOH.

d)half-way between the initial point and the 1^st equivalence point.

e)1.00 mL before you reach the 1^st equivalence point.

f)after the addition of 2.00 mL of KOH past the 1^st equivalence point.

g)at the 2^nd equivalence point.

h)half-way between the 1^st equivalence point and the 2^nd equivalence point.

i)2.00 mL before you reach the 2^nd equivalence point.

j)after the addition of 5.00 mL of KOH past the 2^nd equivalence point.

Answer 1

insert values into the K_a expression for mercaptoacetic acid .

The K_a for mercaptoacetic acid is =1.48x10^-4

HSCH2CO2H -----> HSCH2CO2- + H+

1) at initial 0.00 ml KOH

Ka = [HSCH2CO2-] [H+] / [HSCH2CO2H]

1.48 x 10^-4 = [ (X) (X) / [0.2M]

[H+] = 5.439 x 10^-3 M

pH = -log[H+]

pH = 2.26

2) at the 1st equivalence point.

The solution is now completely composed of a salt of a weak acid. The pH of this solution will be basic.

1) Calculate molarity of KOH

molarity of KOH = 0.12 M

number of moles of acid will equal to base at equivalent point

2.0 x 10^-3 mol / L = 0.12 M

L = 16.66 ml

Calculate the K_b of sodium acetate:

K_w = K_aK_b

1.00 x 10^-14 = (1.48 x 10^-4 ) (x)

x = 6.75 x 10^-11

Calculate pH of the solution:

6.75 x 10^-11 = [(x) (x)] /[0.12 M]

x = 2.84 x 10^-6 M (this is the hydroxide ion concentration)

pOH = 5.54

pH = 8.45

similarly all ..

c) Calculate moles of acid and base in solution before reaction:

CH₃COOH: 2.0x 10^-3 mol
KOH: (0.005 L) (0.120 mol/L) = 6.0 x 10^-4 mol

Determine amounts of acid and acetate ion after reaction:

CH₃COOH: 2.0x 10^-3 mol - 6.0 x 10^-4 mol = 1.4 x 10^-3 mol
CH₃COONa: 6.0 x 10^-4 mol

Use Henderson-Hasselbalch equation to determine pH of (now) buffered solution:

pH = 3.83+ log (6.0 x 10^-4 / 1.4 x 10^-3)

pH = 3.462

similarly do all...